proof of equivalent definitions of analytic sets for Polish spaces


Let A be a nonempty subset of a Polish spaceMathworldPlanetmath X. Then, letting 𝒩 denote Baire spacePlanetmathPlanetmath and Y be any uncountable Polish space, we show that the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    A is -analyticPlanetmathPlanetmath (http://planetmath.org/AnalyticSet2).

  2. 2.

    A is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a closed subset of X×𝒩 onto X.

  3. 3.

    A is the image (http://planetmath.org/DirectImage) of a continuous functionMathworldPlanetmathPlanetmath f:ZX for some Polish space Z.

  4. 4.

    A is the image of a continuous function f:𝒩X.

  5. 5.

    A is the image of a Borel measurable function f:YX.

  6. 6.

    A is the projection of a Borel subset of X×Y onto X.

(1) implies (2): Let be the paving consisting of closed subsets of X. The collectionMathworldPlanetmath of 𝒦-analytic setsMathworldPlanetmath contains the Borel σ-algebra of X (see countable unions and intersections of analytic sets are analytic) and, as the analytic sets are given by a closure operatorPlanetmathPlanetmathPlanetmath (http://planetmath.org/AnalyticSetsDefineAClosureOperator) it follows that it contains all analytic subsets of X. So, any analytic subset A of X is -analytic. Then, there is a closed S𝒩 and a function θ: such that

A=sSn=1θ(sn)

(see proof of equivalent definitions of analytic sets for paved spaces). For each m,n let Km,n denote the closed subset of s𝒩 with sn=m. Then, we can rearrange the above expression to get A=πX(B) where πX:X×𝒩X is the projection map and

B=(X×S)n=1m=1θ(m)×Km,n.

It is easily seen that mθ(m)×Km,n is a closed subset of X×𝒩 for each n, and therefore B is closed, as required.

(2) implies (3): Suppose that A=πX(S) for a closed subset S of X×𝒩, where πX:X×𝒩X is the projection map. As the productPlanetmathPlanetmathPlanetmath of Polish spaces is Polish, and every closed subset of a Polish space is Polish, then S will be a Polish space under the subspace topology. So, we can take Z=S and let f:ZX be the restrictionPlanetmathPlanetmath of πX to Z.

(3) implies (4): Suppose that A is the image of a continuous function g:ZX, for a Polish space Z. As Baire space is universal for Polish spaces, there exists a continuous and onto (http://planetmath.org/SurjectivePlanetmathPlanetmath) function h:𝒩Z. The result follows by taking f=gh.

(4) implies (5): Suppose that A is the image of a continuous function g:𝒩X. Since uncountable Polish spaces are all Borel isomorphic (see Polish spaces up to Borel isomorphism), there is a Borel isomorphism h:Y𝒩. The result follows by taking f=gh.

(5) implies (6): Suppose that A is the image of a Borel measurable function f:YX, and let Γ be its graph (http://planetmath.org/Graph2)

Γ{(f(y),y):yY}X×Y.

The projection of Γ onto X is equal to f(Y)=A, so the result will follow once it is shown that Γ is a Borel set.

Choose a countableMathworldPlanetmath and dense subset {x1,x2,} of X, and let d be a metric generating the topologyMathworldPlanetmath on X. Then, for integers m,n1, denote the open ballPlanetmathPlanetmath about xm of radius 1/n by Bm,n. Since the xm form a dense set, mBm,n=X for each n. Let us define

Γnm=1Bm,n×f-1(Bm,n)X×Y,

which contains Γ. Furthermore, since f-1(Bm,n) are Borel, Γn are Borel sets. Suppose that (x,y)nΓn. Then, for each n, there is an m such that xBm,n and yf-1(Bm,n). So,

d(x,f(y))d(x,xm)+d(xm,f(y))2/n.

This holds for all n, showing that y=f(x) and so (x,y)Γ. We have shown that Γ=nΓn is Borel, as required.

(6) implies (1): This is an immediate consequence of the result that projections of analytic sets are analytic.

Title proof of equivalent definitions of analytic sets for Polish spaces
Canonical name ProofOfEquivalentDefinitionsOfAnalyticSetsForPolishSpaces
Date of creation 2013-03-22 18:48:48
Last modified on 2013-03-22 18:48:48
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Proof
Classification msc 28A05